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Jun 11 What is i^i?

In our last post, we discussed raising a real number to a complex power.

MathJax TeX Test Page We have defined $e^z$ (therefore $n^z$), however we now want to define $z^n$, where z is a complex number. $$z^n = e^{n\log(z)}$$ However, we immediately run into a problem. We have not defined complex logarithms.

Complex Logarithms

MathJax TeX Test Page We begin by writing the equation $$w = \log(z) \to e^w = z$$ We can rewrite w as $u + iv$ and $z$ as $re^{i\theta}$ $$e^{u + iv} = re^{i\theta}$$ So, $e^u = r$ and $e^{iv} = e^{i\theta}$ We now can express u and v. $$u = \ln(r), v = \theta + 2k\pi$$ Now, we combine our answer to get the logarithm. $$w = \ln(r) + i\left(\theta + 2k\pi\right)$$

Quick Example

MathJax TeX Test Page $$\ln(-1)$$ Well, we have to figure out two things: the modulus (magnitude) of this complex number and its argument (angle). The magnitude is $1$, and the angle is $\pi$. So, the logarithm equals $$\ln(1) + i\left(\pi + 2k\pi\right)$$ $$i\left(\pi + 2k\pi\right)$$

So, what is i^i?

MathJax TeX Test Page $$i^i = e^{i\ln(i)}$$ $$e^{i\ln(i)} = e^{i*\left(i\frac{\pi}{2}\right)}$$ $$= e^{-\frac{\pi}{2}} = \boxed{0.207}$$ That is a pretty crazy result, because we raised a complex number to a complex power and got a real number. Another way to think about this is $$i = e^{i\frac{\pi}{2}} \to i^i = e^{i\frac{\pi}{2} \cdot i}$$ Now, we get the same result, $e^{-\frac{\pi}{2}}$
David Witten